Download presentation

Presentation is loading. Please wait.

Published byJulian Morrison Modified over 6 years ago

1
3.3: Dividing Polynomials: Remainder and Factor Theorems Long Division of Polynomials 1.Arrange the terms of both the dividend and the divisor in descending powers of any variable. 2.Divide the first term in the dividend by the first term in the divisor. The result is the first term of the quotient. 3.Multiply every term in the divisor by the first term in the quotient. Write the resulting product beneath the dividend with like terms lined up. 4.Subtract the product from the dividend. 5.Bring down the next term in the original dividend and write it next to the remainder to form a new dividend. 6.Use this new expression as the dividend and repeat this process until the remainder can no longer be divided. This will occur when the degree of the remainder (the highest exponent on a variable in the remainder) is less than the degree of the divisor. 1.Arrange the terms of both the dividend and the divisor in descending powers of any variable. 2.Divide the first term in the dividend by the first term in the divisor. The result is the first term of the quotient. 3.Multiply every term in the divisor by the first term in the quotient. Write the resulting product beneath the dividend with like terms lined up. 4.Subtract the product from the dividend. 5.Bring down the next term in the original dividend and write it next to the remainder to form a new dividend. 6.Use this new expression as the dividend and repeat this process until the remainder can no longer be divided. This will occur when the degree of the remainder (the highest exponent on a variable in the remainder) is less than the degree of the divisor.

2
Example:Long Division of Polynomials Divide 4 – 5x – x 2 + 6x 3 by 3x – 2. Solution We begin by writing the divisor and dividend in descending powers of x. Next, we consider how many times 3x divides into 6x 3. 3x – 2 6x 3 – x 2 – 5x + 4 2x22x2 Divide: 6x 3 /3x = 2x 2. 6x 3 – 4x 2 Multiply. Multiply: 2x 2 (3x – 2) = 6x 3 – 4x 2. Divide: 3x 2 /3x = x. Now we divide 3x 2 by 3x to obtain x, multiply x and the divisor, and subtract. 3x – 2 6x 3 – x 2 – 5x + 4 6x 3 – 4x 2 2x 2 + x 3x 2 – 5x Multiply. Multiply: x(3x – 2) = 3x 2 – 2x. 3x 2 – 2x more Subtract 3x 2 – 2x from 3x 2 – 5x and bring down 4. + 4 -3x – 5x Subtract 6x 3 – 4x 2 from 6x 3 – x 2 and bring down –5x. 3x23x2 3.3: Dividing Polynomials: Remainder and Factor Theorems

3
Example:Long Division of Polynomials Divide 4 – 5x – x 2 + 6x 3 by 3x – 2. Divide: -3x/3x = -1. Solution Now we divide –3x by 3x to obtain –1, multiply –1 and the divisor, and subtract. 3x – 2 6x 3 – x 2 – 5x + 4 6x 3 – 4x 2 2x 2 + x – 1 3x 2 – 5x + 4 3x 2 – 2x -3x Multiply. Multiply: -1(3x – 2) = -3x + 2. -3x + 2 Subtract -3x + 2 from -3x + 4, leaving a remainder of 2. 2 3.3: Dividing Polynomials: Remainder and Factor Theorems

4
The Division Algorithm If f (x) and d(x) are polynomials, with d(x) = 0, and the degree of d(x) is less than or equal to the degree of f (x), then there exist unique polynomials q(x) and r(x) such that f (x) = d(x) q(x) + r(x). The remainder, r(x), equals 0 or its is of degree less than the degree of d(x). If r(x) = 0, we say that d(x) divides evenly in to f (x) and that d(x) and q(x) are factors of f (x). If f (x) and d(x) are polynomials, with d(x) = 0, and the degree of d(x) is less than or equal to the degree of f (x), then there exist unique polynomials q(x) and r(x) such that f (x) = d(x) q(x) + r(x). The remainder, r(x), equals 0 or its is of degree less than the degree of d(x). If r(x) = 0, we say that d(x) divides evenly in to f (x) and that d(x) and q(x) are factors of f (x). Dividend Divisor Quotient Remainder 3.3: Dividing Polynomials: Remainder and Factor Theorems

5
Synthetic Division To divide a polynomial by x – c Example 1. Arrange polynomials in descending powers, with a 0 coefficient for any missing terms. x – 3 x 3 + 4x 2 – 5x + 5 2. Write c for the divisor, x – c. To the right, 3 1 4 -5 5 write the coefficients of the dividend. 3. Write the leading coefficient of the dividend 3 1 4 -5 5 on the bottom row. Bring down 1. 1 4. Multiply c (in this case, 3) times the value 3 1 4 -5 5 just written on the bottom row. Write the 3 product in the next column in the 2nd row. 1 To divide a polynomial by x – c Example 1. Arrange polynomials in descending powers, with a 0 coefficient for any missing terms. x – 3 x 3 + 4x 2 – 5x + 5 2. Write c for the divisor, x – c. To the right, 3 1 4 -5 5 write the coefficients of the dividend. 3. Write the leading coefficient of the dividend 3 1 4 -5 5 on the bottom row. Bring down 1. 1 4. Multiply c (in this case, 3) times the value 3 1 4 -5 5 just written on the bottom row. Write the 3 product in the next column in the 2nd row. 1 Multiply by 3. 3.3: Dividing Polynomials: Remainder and Factor Theorems

6
Synthetic Division 5. Add the values in this new column, writing the sum in the bottom row. 6. Repeat this series of multiplications and additions until all columns are filled in. 7. Use the numbers in the last row to write the quotient and remainder in fractional form. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in the row is the remainder. 5. Add the values in this new column, writing the sum in the bottom row. 6. Repeat this series of multiplications and additions until all columns are filled in. 7. Use the numbers in the last row to write the quotient and remainder in fractional form. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in the row is the remainder. 314-55 3 17 Add. 314-55 32148 171653 Add. Multiply by 3. 314-55 321 1716 Add. Multiply by 3. 1x 2 + 7x + 16 + 53 x – 3 3.3: Dividing Polynomials: Remainder and Factor Theorems

7
Example:Using Synthetic Division Use synthetic division to divide 5x 3 + 6x + 8 by x + 2. Solution The divisor must be in the form x – c. Thus, we write x + 2 as x – (-2). This means that c = -2. Writing a 0 coefficient for the missing x 2 - term in the dividend, we can express the division as follows: x – (-2) 5x 3 + 0x 2 + 6x + 8. more Now we are ready to set up the problem so that we can use synthetic division. -25068 Use the coefficients of the dividend in descending powers of x. This is c in x-(-2). 3.3: Dividing Polynomials: Remainder and Factor Theorems

8
Solution We begin the synthetic division process by bringing down 5. This is following by a series of multiplications and additions. 5. Add: 6 + 20 = 26. -25068 -1020 5-1026 Add. 7. Add: 8 + (-52) = -44. -25068 -1020-52 5-1026-44 Add. 4. Multiply: -2(-10) = 20. -25068 -1020 5-10 6. Multiply: -2(26) = -52. -25068 -1020-52 5-1026 more -25068 5 1. Bring down 5. 3. Add: 0 + (-10) = -10. -25068 -10 5-10 Add. 2. Multiply: -2(5) = -10. -25068 -10 5 3.3: Dividing Polynomials: Remainder and Factor Theorems

9
Solution -25068 -1020-52 5-1026-44 The numbers in the last row represent the coefficients of the quotient and the remainder. The degree of the first term of the quotient is one less than that of the dividend. Because the degree of the dividend is 3, the degree of the quotient is 2. This means that the 5 in the last row represents 5x 2. Thus, x + 2 5x 3 + 0x 2 + 6x + 8 5x 2 – 10x + 26 – x + 2 44 3.3: Dividing Polynomials: Remainder and Factor Theorems

10
The Remainder Theorem If the polynomial f (x) is divided by x – c, then the remainder is f (c). 3.3: Dividing Polynomials: Remainder and Factor Theorems

11
The Factor Theorem Let f (x) be a polynomial. 1.If f (c ) = 0, then x – c is a factor of f (x). 2.If x – c is a factor of f (x), then f ( c) = 0. Let f (x) be a polynomial. 1.If f (c ) = 0, then x – c is a factor of f (x). 2.If x – c is a factor of f (x), then f ( c) = 0. 3.3: Dividing Polynomials: Remainder and Factor Theorems

12
Example:Using the Factor Theorem Solve the equation 2x 3 – 3x 2 – 11x + 6 = 0 given that 3 is a zero of f (x) = 2x 3 – 3x 2 – 11x + 6. Solution We are given that f (3) = 0. The Factor Theorem tells us that x – 3 is a factor of f (x). We’ll use synthetic division to divide f (x) by x – 3. Equivalently, 2x 3 – 3x 2 – 11x + 6 = (x – 3)(2x 2 + 3x – 2) 32-3-116 69-6 23-20 2x 3 – 3x 2 – 11x + 6 x – 3 2x 2 + 3x – 2 more 3.3: Dividing Polynomials: Remainder and Factor Theorems

13
Solution Now we can solve the polynomial equation. The solution set is {-2, ½, 3}. 2x 3 – 3x 2 – 11x + 6 = 0 This is the given equation. (x – 3)(2x 2 + 3x – 2) = 0 Factor using the result from the synthetic division. (x – 3)(2x – 1)(x + 2) = 0 Factor the trinomial. x – 3 = 0 or 2x – 1 = 0 or x + 2 = 0 Set each factor equal to 0. x = 3 x = ½ x = -2 Solve for x. 3.3: Dividing Polynomials: Remainder and Factor Theorems

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google